Which of the definite integrals is equivalent to the following limit? $ \lim_{n\to\infty} \sum_{i=1}^n \sqrt{\dfrac{9(i-1)}n}\cdot\dfrac9n$ Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_0^9 \sqrt {9x}\,dx$ (Choice B) B $ \int_1^9 \sqrt {9x}\,dx$ (Choice C) C $ \int_1^9 \sqrt x\,dx$ (Choice D) D $ \int_0^9 \sqrt x\,dx$
Explanation: The value of a definite integral is the limit of its Riemann sums as the number of terms tends to infinity. The given summation $ \sum_{i=1}^n \sqrt{\dfrac{9(i-1)}n}\cdot\dfrac9n$ looks like a left Riemann sum with $n$ subintervals of equal width. If each subinterval has width $\Delta x$, what is the left Riemann sum for the following definite integral? $ \int_a^b f(x) \,dx$ The left Riemann sum for the definite integral is $ \sum_{i=1}^n f(a+(i-1)\Delta x)\cdot\Delta x$. What does the sum become when we express $\Delta x$ in terms of $a$, $b$, and $n$ ? Dividing the interval $[a,b]$ into $n$ subintervals of equal width yields a common width of $\Delta x=\dfrac{b-a}n\,$. This lets us express the left Riemann sum as $ \sum_{i=1}^n f\left(a+(i-1)\cdot\dfrac{b-a}n\right)\cdot\dfrac{b-a}n$. Let's rewrite the given summation as $ \sum_{i=1}^n \sqrt{(i-1)\cdot\dfrac9n}\cdot\dfrac9n$ If the given summation is a left Riemann sum, what are $a$, $b$, and $f$ ? Equating the width $\Delta x$ of each subinterval in the two sums yields $\dfrac{b-a}n=\dfrac9n\,$. Thus, the interval $[a,b]$ has width $b-a=9$. In two of the answer choices, $f(x)=\sqrt x$. In that case, $a=0$ and $b=a+9=9$. These values produce the definite integral $ \int_0^9 \sqrt x \,dx$. In two other answer choices, $f(x)=\sqrt{9x}$. To accommodate this function, we need to move the constants around to rewrite the given summation as $ 9\,\sum_{i=1}^n \sqrt{9(i-1)\cdot\dfrac1n}\cdot\dfrac1n\,$. The summation is now a left Riemann sum with $\Delta x=1/n$, which means $b-a=1$. We have $a=0$ and $b=a+1=1$. These values produce the definite integral $ 9\int_0^1 \sqrt{9x} \,dx$. Are either of these two integral expressions answer choices? The correct answer is $ \lim_{n\to\infty} \sum_{i=1}^n \sqrt{\dfrac{9(i-1)}n} \cdot\dfrac9n = \int_0^9 \sqrt x \,dx$.